If the diagonals of a parallelogram are represented by the vectors ` 3hati + hatj -2hatk and hati + 3hatj -4hatk`, then its area in square units , is

Options

A. ` 5sqrt3`
B. `6sqrt3`
C. `sqrt42`
D. `sqrt42`

Explanation

To find the area of the parallelogram given its diagonals represented by the vectors \( \mathbf{d_1} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \mathbf{d_2} = \hat{i} + 3\hat{j} - 4\hat{k} \), we can use the formula for the area of a parallelogram in terms of its diagonals:

\[ \text{Area} = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| \]

### Step 1: Calculate the Cross Product \( \mathbf{d_1} \times \mathbf{d_2} \)

We set up the determinant for the cross product:

\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 3 & -4 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Using the determinant formula, we expand it as follows:

\[ \mathbf{d_1} \times \mathbf{d_2} = \hat{i} \begin{vmatrix} 1 & -2 \\ 3 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -2 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (-2)(3) = -4 + 6 = 2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} = (3)(-4) - (-2)(1) = -12 + 2 = -10 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} = (3)(3) - (1)(1) = 9 - 1 = 8 \]

### Step 3: Combine the Results

Now substituting back into the cross product:

\[ \mathbf{d_1} \times \mathbf{d_2} = 2\hat{i} + 10\hat{j} + 8\hat{k} \]

### Step 4: Calculate the Magnitude of the Cross Product

Now we find the magnitude:

\[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{(2)^2 + (10)^2 + (8)^2} = \sqrt{4 + 100 + 64} = \sqrt{168} \]

### Step 5: Calculate the Area of the Parallelogram

Finally, we calculate the area:

\[ \text{Area} = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \frac{1}{2} \sqrt{168} = \frac{1}{2} \cdot 2\sqrt{42} = \sqrt{42} \]

Thus, the area of the parallelogram is:

\[ \text{Area} = \sqrt{42} \text{ square units} \]

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